Two enterprising chemistry students decide to develop an alternative to Alka-Sel

Question

Two enterprising chemistry students decide to develop an alternative to Alka-Seltzer tablets, replacing the aspirin (acetylsalicylic acid) with vitamin c (ascorbic acid). The following table shows one of the formulations they tried for a 5 g tablet.

When the tablet dissolves in water, it "frizzes" as the bicarbonate reacts with the citric acid and the vitamin C. What mass of NaHCO3 reacts with the vitamin C?

Citric acid is triprotic, and reacts with the base to form the citrate ino (C6H5O73-). Write the overall ionic reaction for the acidification of the NaHCO3 by the citric acid, giving CO2 as the product. Citric acid should be written as H3C6H5O7 in the net ionic equation, as weak acids are written in the molecular form even though they slightly ionize.

What mass of the NaHCO3 reacts with the citric acid?

What is the mass of CO2 released in this reaction?

THe remaining NaHCO3 is available to neautralize stomach acid. What mass of stomach HCL could be neutralized by one tablet?

ingredient formula Quantity (mg) Sodium bicarbonate NaHCO3 1991 citric acid H3C6H5O7 886 vitamin c HC6H7O6 370 inert binder 1753 TOTAL MASS: 5000

Explanation / Answer

What mass of NaHCO3 reacts with the vitamin C?

Since vitamin c is monoprotic acid and so mole ratio between NaHCO3 and vitamin C is 1 :1

Moles of vitamin c = 370 E -3 g / molar mass of vitamin c = 370 E-3 g / 176.12 g per mol = 0.0021 mol

Moles of NaHCO3 = 0.0021 mol NaHCO3 = 0.0021 mol x 80.007 g per mol = 0.168 g = 168 mg

Citric acid is triprotic, and reacts with the base to form the citrate ino (C6H5O73-). Write the overall ionic reaction for the acidification of the NaHCO3 by the citric acid, giving CO2 as the product.

3NaHCO3(s) + H3C6H5O7(aq)   -- > Na3C6H5O7(aq) + 3CO2(g) + 3H2O(l)

Ionic equation :

3NaHCO3(s) + H3C6H5O7(aq)   -- > 3 Na+ (aq) + C6H5O73- (aq) + 3CO2(g) + 3H2O(l)

What mass of the NaHCO3 reacts with the citric acid?

Calculation of mass :

moles of NaHCO3 = (mass of citric acid / molar mass ) x ( 3 mol NaHCO3 / 1 mol citric acid )

= (0.886 g / 192.124 g per mol )x ( 3 mol NaHCO3 / 1 mol Citric acid )

=0.0138 mol NaHCO3

Mass of NaHCO3 = 0.0138 mol x 80.007 g /mol = 1.11 g

What is the mass of CO2 released in this reaction

Moles of CO2 = Moles of NaHCO3 x 3 mol CO2 / 3 mol NaHCO3

= 0.0138 mol CO2

Mass of CO2 = 0.0138 mol x 44.009 g per mol= 0.61 g

THe remaining NaHCO3 is available to neautralize stomach acid. What mass of stomach HCL could be neutralized by one tablet?

Reacted mass of NaHCO3 = 1.11 g

Excess mass of NaHCO3 = 1.991 g - 1.11 g = 0.88 g

Calculation of moles of NaHCO3 = 0.88 g / 84.007 g per mol= 0.0105 mol

Mole ration of reaction of sodium carbonate with HCl is 1 : 1 since HCl is monoprotic acid

Therefore moles of HCl = moles of sodium bicarbonate = 0.0105

Mass of HCl = 0.0105 mol x 36.46 g /mol

= 0.384 g

Mass of HCl = 0.384 g = 384 mg mass is neutralized by NaHCO3

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